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familiari (nf3795) – quantum atomic – yeazell – (58395) 1This print-out should have 17 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsA helium-neon laser emits l ight of wavelength632.8 nm a nd has a power output o f 20 mW.How many photons are emitted per secondby this la ser?Correct a nswer: 6.37903 × 1016.Explanation:Let : P = 20 mW = 0.02 W ,λ = 632.8 nm , andhc = 1240 eV · nm .The energy of one photon isEphoton=hcλ,so the number of photons emitted per secondby this la ser isn =PEphoton=λ Phc=(632.8 nm) (0.02 W)1240 eV · nm1 eV1.6 × 10−19J=6.37903 × 1016.002 10.0 pointsYou do a photoelectric ex periment and youfind two different values of the maximum ki-netic energy of photoelectrons emitted from ametallic surface when light of two frequenciesis incident on the surface.When light of a frequency7.09167 × 1014Hz is used the maximum ki-netic energy is 2 × 10−19J. When light of afrequency 7.65626 × 1014Hz is used the max-imum kinetic energy is 2.4 × 10−19J.From your rough experiment, find a valuefor P lanck’s constant. Hint: It is related tothe slope.Correct answer: 7.08482 × 10−34Js.Explanation:The k inetic energy of the electron is deter-mined by the energy of the photon minus theenergy needed to remove the electron fromthe surface.K = hf − WNote that h is the slope in this function sousing the two points we can find this slopeor this experimental measurement of Planck’sconstant, 7.08482 × 10−34m/s.003 10.0 pointsOne electron travels twice a s fast as another.Which has longer wavelength?1. The slower one correct2. More information is needed.3. The two have the same wavelength.4. The faster oneExplanation:The twice-as-fast electron has twice the mo-mentum. By de Broglie’s formula, wavelengthis inversely proportional to momentum, sotwice the momentum means half the wave-length. Thus the slower electron has thelonger wavelength.004 10.0 pointsWhat is the uncertainty in the speed of aproton if it is desired to l ocate its positionto within an uncertainty of x = 1.91 nm ?Planck’s constant is 6.63 × 10−34J · s and themass of the proton is 1.67 × 10−27kg.Correct answer: 16.5407 m/s.Explanation:Let : h = 6.63 × 10−34J · s ,m = 1.67 × 10−27kg , and∆x = 1.91 nm = 1.91 × 10−9m .familiari (nf3795) – quantum atomic – yeazell – (583 95) 2From the H eisenberg uncertainty principle,∆x ∆px≥h4 π∆x m ∆vx≥h4 π∆vx≥h4 π m ∆x=6.63 × 10−34J · s4 π (1.67 × 10−27kg)×11.91 × 10−9m=16.5407 m/s .005 (part 1 of 2) 10.0 pointsA hypothetical atom has four energy states asshown.00−1−2−3−4−5−6−7−8n = 4n = 3n = 2n = 1Energy (eV)Which of the following photon energies Eγcould NOT be found in the emission spectraof this atom after it has been excited to then = 4 state?1. Eγ= 1 eV2. Eγ= 5 eV3. Eγ= 4 eV correct4. Eγ= 2 eV5. Eγ= 3 eVExplanation:By subtraction, t he transition from n = 4to n = 3 would release 1 eV; from n = 4to n = 2 would release 3 eV; from n = 3 ton = 2 would release 2 eV; from n = 3 to n = 1would release 5 eV. The only choice remainingis4 eV .006 (part 2 of 2) 10.0 pointsWhich of the followi ng transitions will pro-duce the photon with the longest wavelength?1. n = 4 to n = 3 correct2. n = 4 to n = 13. n = 3 to n = 24. n = 3 to n = 15. n = 2 to n = 1Explanation:The longest wavelength is pro duced inthe case where the least a mount of energyis released. That is, in this case, fromn = 4 to n = 3 .007 (part 1 of 3) 10.0 pointsConsider just four of the energy levels in acertain a tom, as shown in the diagram below.n = 1n = 2n = 3n = 4How many spectral lines will result fromall possible transitions among these levels?Which transition correspo nds to t he highest-frequency light emi tted? Which transitioncorresponds to the lowest-frequency?1. three; level 4 to level 3 transition; level 2to level 1 transition.familiari (nf3795) – quantum atomic – yeazell – (583 95) 32. three; l evel 2 t o level 1 transition; level 4to level 3 transition.3. six; level 4 t o level 1 transition; level 4 tolevel 3 transit ion. correct4. three; l evel 4 t o level 1 transition; level 4to level 3 transition.Explanation:Six transitions are possible, as shown.n = 1n = 2n = 3n = 4The highest-frequency transition is fromquantum level 4 to level 1. The lowest-frequency transition is from quantum level4 to level 3.008 (part 2 of 3) 10.0 pointsAn electron de-excites from the fourth q uan-tum level t o the third and then di rectly to theground state. Two photo ns ar e emitted.How does the sum of their frequencies com-pare to the frequency of the single photonthat would be emitted by de-excitation fromthe fourth level dir ectly to the ground state?1. The sum is larger than the frequency ofthe single photon.2. The sum is smaller than the frequency ofthe single photon.3. None o f these4. The sum is equal to the frequency of thesingle photon. correctExplanation:As shown in the diag ram, the sum of thetwo frequencies is equal to the frequency oflight emitted in the transition from q uantumlevel 4 to the ground state, quantum level1. Because energies are additive, so are thefrequencies. However, the wavelength rela-tionship is not that simple: wavelength is in-versely proportional to the frequency, so thereciprocals of the wavelengths are additive:1λ4→3+1λ3→1=1λ4→1009 (part 3 of 3) 10.0 pointsSuppose the four energy levels were somehowevenly spaced.How many spectral lines woul d result?1. six2. five3. four4. three correctExplanation:Only three.n = 1n = 2n = 3n = 4The transit ion from 4 to 3 would involvethe same difference in energy and be indis-tinguishable from the transition from 3 to 2,or from 2 to gro und. Likewise, the transitionfrom 4 to 2 would have the same change inenergy as a transition from 3 to ground.010 10.0 pointsA 6. 5 µg particl e is moving with a speed ofapproximately 0.2 cm/s in a box of length1.514 cm.familiari (nf3795) – quantum atomic – yeazell – (583 95) 4Treating this as a one-dimensional particlein a box , calculat e the a pproximate value ofthe quantum number n.Correct a nswer: 5.93725 × 1020.Explanation:Let : h = 6.63 × 10−34J · s ,m = 6.5 µg = 6.5 × 10−9kg ,v = 0.2 cm/s = 0.002 m/s , andL = 1 .514 cm = 0.01514 m .The energy of the particl e when it is in thenth st a te isEn=n2h28 m L2=m v22, son =2 m v Lh=2 (6.5 × 10−9kg) (0.002 m/s) (0.01514 m)6.63 × 10−34J · s=5.93725 ×


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